Show that the three qubit gate $G$ defined by the circuit is universal for quantum computation whenever $\alpha$ is irrational.
My Observations
The unitary gate on the third qubit is activated only when the first two qubits are both in the state $1$.
$$ |\psi_{in}\rangle=|\psi_{1}\rangle\otimes|\psi_{2}\rangle\otimes|\psi\rangle=(a|0\rangle+b|1\rangle)\otimes(c|0\rangle+d|1\rangle)\otimes|\psi\rangle=(ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle)\otimes|\psi\rangle\\ =ac|00\rangle\otimes|\psi\rangle+ad|01\rangle\otimes|\psi\rangle+bc|10\rangle\otimes|\psi\rangle+bd|11\rangle\otimes|\psi\rangle $$ $$ |\psi_{out}\rangle=ac|00\rangle\otimes|\psi\rangle+ad|01\rangle\otimes|\psi\rangle+bc|10\rangle\otimes|\psi\rangle+bd|11\rangle\otimes iR_x(\pi \alpha)|\psi\rangle $$
We can also prove that arbitrarily fine approximations of all angles can be achieved using a rotation by an irrational multiple of $\pi$.
I have no clue on how this circuit can be used for universal quantum gate construction ?
Refer to Exercise 4.44, Page198, Quantum Computation and Quantum Information by Nielsen and Chuang
Thanks @Egretta.Thula for the link where @DaftWullie's answer says:
Firstly, imagine you start from a two-qubit state $|00⟩$, and apply an $R_x$ rotation with an angle equivalent to half that of a Pauli $X$ to the first qubit (I forget which convention N&C is using for their rotation gates). Then apply a controlled-not controlled off the first qubit and targeting the second qubit. Next, apply the inverse of the first rotation. Finally, measure the first qubit. If you get answer $|1⟩$, the second qubit is in the $|−⟩$ state. If it isn't, discard and repeat. So, we can produce the $|−⟩$ state. If you input this as the target qubit of the controlled-controlled-$R_x$ (of arbitrary rotation angle), and have one of the controls in the $|1⟩$ state, you get an arbitrary $Z$ rotation on the other control qubit.
So, we know we can do arbitrary $X$ and $Z$ rotations, meaning that you can make any single-qubit unitary. Combine that with controlled-not and you know you have universality.
This is not with any diagrams nor any equations and I do not get a clue of what it is suggesting to do ?
My Attempt
Imagine you start from a two-qubit state $|00⟩$, and apply an $R_x$ rotation with an angle equivalent to half that of a Pauli $X$ to the first qubit (I forget which convention N&C is using for their rotation gates). Then apply a controlled-not controlled off the first qubit and targeting the second qubit. Next, apply the inverse of the first rotation. Finally, measure the first qubit. If you get answer $|1⟩$, the second qubit is in the $|−⟩$ state $$ M|00\rangle=\frac{i}{\sqrt{2}}\begin{bmatrix}1&-i\\-i&1\end{bmatrix}|00\rangle=\frac{1}{\sqrt{2}}[i|0\rangle+|1\rangle]\otimes|0\rangle=\frac{1}{\sqrt{2}}[i|00\rangle+|10\rangle] $$ $$ M^{-1}[\frac{1}{\sqrt{2}}[i|00\rangle+|10\rangle]]=\frac{1}{2}[i(-i|0\rangle+|1\rangle)\otimes |0\rangle+(|0\rangle-i|1\rangle)\otimes|1\rangle]\\ =\frac{1}{2}[|00\rangle+i|10\rangle+|01\rangle-i|11\rangle]=\frac{1}{\sqrt{2}}[|0\rangle(\frac{|0\rangle+|1\rangle}{\sqrt{2}})+i|1\rangle(\frac{|0\rangle-|1\rangle}{\sqrt{2}})]\\ =\frac{1}{\sqrt{2}}[|0\rangle|+\rangle+i|1\rangle|-\rangle] $$
